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Neglecting gravity, the equation of motion is $m\ddot{\mathbf{x}}=a\dot{\mathbf{x}}\times \mathbf{x}/r^3$, where $r=|\mathbf{x}|>0$. Dotting with the velocity gives $d(m|\dot{\mathbf{x}}{|}^2/2)/dt=0$, because a cross product is perpendicular to its factors. Thus the speed is constant. The vector triple product givesHence the generalized angular momentum in a radial monopole field isDotting with $\mathbf{x}$ gives $\mathbf{L}\cdot \mathbf{x}=-ar$, or $\hat{\mathbf{L}}\cdot \hat{\mathbf{x}}=-a/|\mathbf{L}|$. For $\mathbf{L}\ne 0$, the direction of $\mathbf{x}$ therefore makes the fixed angle $\boxed{\cos \beta =-a/|\mathbf{L}|}$ with the axis $\mathbf{L}$, describing a right circular conical surface with vertex at the North Pole. Orthogonality gives $|\mathbf{L}{|}^2=m^2|\mathbf{x}\times \dot{\mathbf{x}}{|}^2+a^2$, so this cosine is in $[-1,1]$. Radial motion gives a degenerate cone; when $a=0$ and the ordinary angular momentum is nonzero the cone is the usual orbital plane. If $\mathbf{L}=0$, the norm identity forces $a=0$ and $\mathbf{x}\times \dot{\mathbf{x}}=0$, giving a radial straight line.

For $\psi =\mu /r$, the radial gradient givesThe force $-\nabla \varphi$ is radial and has no torque about the origin. For the remaining force, use the vector triple product:Meanwhile the radial unit vector obeysSince $\psi \mathbf{r}=\mu \hat{\mathbf{r}}$, these derivatives cancel. Thus the generalized angular momentum in a radial monopole field isThe argument applies away from $r=0$, where the field is defined, for any radial $\varphi$. In contrast to ordinary angular momentum, this conserved vector has $\mathbf{L}\cdot \hat{\mathbf{r}}=-\mu$; it fixes a cone rather than generally fixing a plane.

This surface consists of rays from a vertex $\mathbf{a}$ whose directions make a fixed angle $\beta$ with a unit axis vector $\hat{\mathbf{n}}$. The displayed equation follows from the dot product formula for that angle. For $0<\beta <\pi$, $\beta \ne \pi /2$, it gives one circular nappe; $\beta =\pi /2$ gives a plane, and $\beta =0,\pi$ gives a degenerate ray. Squaring the equation produces both nappes, so the unsquared sign distinguishes them. Generalized angular momentum in a radial monopole field confines particle directions to such a surface.